1.
Calculate the number of protons, neutrons, and electrons in $^{35}_{17}\mathrm{Cl}$ and $^{80}_{35}\mathrm{Br}$.
Solution:
For $^{35}_{17}\mathrm{Cl}$: Atomic number (Z) = 17, Mass number (A) = 35.
Protons = Z = 17. Electrons = Z = 17 (neutral atom). Neutrons = A - Z = $35 - 17 = 18$.
For $^{80}_{35}\mathrm{Br}$: Atomic number (Z) = 35, Mass number (A) = 80.
Protons = Z = 35. Electrons = Z = 35. Neutrons = A - Z = $80 - 35 = 45$.
2.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the mass number of the element.
Solution:
(i) For a neutral atom, number of protons = number of electrons. Therefore, Protons = 29.
(ii) Mass number (A) = Number of protons + Number of neutrons = $29 + 35 = 64$.
3.
In Rutherford's scattering experiment, which subatomic particle was used to bombard the thin gold foil, and what conclusion was drawn from the observation that a very small fraction of particles bounced back?
Solution:
Particle used: Alpha ($\alpha$) particles (Doubly charged Helium ions, $\mathrm{He^{2+}}$).
Conclusion: The fraction bouncing back (deflection by 180°) proved that the entire positive charge and almost all the mass of the atom is concentrated in a very small, dense, central volume called the nucleus.
4.
Calculate the mass and charge of one mole of electrons.
Solution:
Mass: Mass of 1 e⁻ = $9.109 \times 10^{-31}$ kg.
Mass of 1 mole = $(9.109 \times 10^{-31}) \times (6.022 \times 10^{23}) \approx 5.48 \times 10^{-7}$ kg.
Charge: Charge of 1 e⁻ = $1.602 \times 10^{-19}$ C.
Charge of 1 mole = $(1.602 \times 10^{-19}) \times (6.022 \times 10^{23}) \approx 96485$ C (which is defined as 1 Faraday).
5.
Arrange the fundamental particles (electron, proton, neutron, alpha particle) in increasing order of their specific charge (e/m ratio).
Solution:
Specific charge = charge(e) / mass(m).
1. Neutron: charge = 0 $\Rightarrow$ e/m = 0.
2. Alpha ($\alpha$) particle: charge = $+2e$, mass = $\approx 4m_p$ $\Rightarrow$ e/m $\approx 2/4 = 0.5$ (relative).
3. Proton: charge = $+1e$, mass = $m_p$ $\Rightarrow$ e/m = 1 (relative).
4. Electron: charge = $-1e$, mass = negligible ($m_p/1837$) $\Rightarrow$ e/m is extremely high.
Order: Neutron $<$ Alpha particle $<$ Proton $<$ Electron.
6.
Calculate the wavelength, frequency, and wavenumber of a light wave whose period is $2.0 \times 10^{-10}$ s.
Solution:
Period ($T$) = $2.0 \times 10^{-10}$ s.
Frequency ($\nu$): $\nu = \frac{1}{T} = \frac{1}{2.0 \times 10^{-10}} = 5.0 \times 10^9 \text{ s}^{-1}$ (or Hz).
Wavelength ($\lambda$): $\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{5.0 \times 10^9} = 0.06 \text{ m}$.
Wavenumber ($\bar{\nu}$): $\bar{\nu} = \frac{1}{\lambda} = \frac{1}{0.06} = 16.66 \text{ m}^{-1}$.
7.
Find the energy of a photon which corresponds to light of frequency $3 \times 10^{15}$ Hz. (Given $h = 6.626 \times 10^{-34}$ J s)
Solution:
Energy ($E$) = $h\nu$
$E = (6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^{15} \text{ s}^{-1})$
$E = 19.878 \times 10^{-19} \text{ J} = 1.988 \times 10^{-18} \text{ J}$.
8.
Calculate the energy of one mole of photons of radiation whose frequency is $5 \times 10^{14}$ Hz.
Solution:
Energy of 1 photon ($E$) = $h\nu = (6.626 \times 10^{-34}) \times (5 \times 10^{14}) = 3.313 \times 10^{-19} \text{ J}$.
Energy of 1 mole of photons ($E_{mol}$) = $E \times N_A$
$E_{mol} = (3.313 \times 10^{-19}) \times (6.022 \times 10^{23}) \approx 199.5 \times 10^3 \text{ J} = 199.5 \text{ kJ mol}^{-1}$.
9.
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Solution:
Power = 100 W = 100 J/s. (Total energy emitted per second).
Energy of one photon ($E$) = $\frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} = 4.969 \times 10^{-19} \text{ J}$.
Number of photons emitted per second ($n$) = $\frac{\text{Total Energy}}{\text{Energy of 1 photon}}$
$n = \frac{100}{4.969 \times 10^{-19}} = 2.012 \times 10^{20} \text{ photons/sec}$.
10.
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter.
Solution:
Frequency ($\nu$) = $1,368 \text{ kHz} = 1368 \times 10^3 \text{ Hz}$.
Wavelength ($\lambda$) = $\frac{c}{\nu}$
$\lambda = \frac{3 \times 10^8}{1368 \times 10^3} \approx 219.3 \text{ m}$.
11.
The threshold frequency ($\nu_0$) for a metal is $7.0 \times 10^{14}$ s$^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $\nu = 1.0 \times 10^{15}$ s$^{-1}$ hits the metal.
Solution:
Einstein's Photoelectric Equation: $K.E. = h(\nu - \nu_0)$
$K.E. = 6.626 \times 10^{-34} \times (1.0 \times 10^{15} - 0.7 \times 10^{15})$
$K.E. = 6.626 \times 10^{-34} \times (0.3 \times 10^{15}) = 1.988 \times 10^{-19} \text{ J}$.
12.
A photon of wavelength $4 \times 10^{-7}$ m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron.
Solution:
(i) Energy of photon: $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.969 \times 10^{-19} \text{ J}$.
In eV: $E = \frac{4.969 \times 10^{-19}}{1.602 \times 10^{-19}} = 3.10 \text{ eV}$.
(ii) $K.E. = E - W_0 = 3.10 \text{ eV} - 2.13 \text{ eV} = 0.97 \text{ eV}$.
(iii) $K.E.$ in Joules = $0.97 \times 1.602 \times 10^{-19} = 1.554 \times 10^{-19} \text{ J}$.
$K.E. = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 1.554 \times 10^{-19}}{9.109 \times 10^{-31}}} \approx 5.84 \times 10^5 \text{ m/s}$.
13.
When light with a wavelength of 400 nm falls on the surface of potassium metal, electrons with a kinetic energy of $0.85$ eV are emitted. Find the work function of potassium in Joules.
Solution:
Energy of incident light: $E = \frac{1240}{\lambda (\text{in nm})} \text{ eV} = \frac{1240}{400} = 3.10 \text{ eV}$.
From equation: $E = W_0 + K.E. \Rightarrow W_0 = E - K.E. = 3.10 - 0.85 = 2.25 \text{ eV}$.
Convert to Joules: $W_0 = 2.25 \times 1.602 \times 10^{-19} = 3.604 \times 10^{-19} \text{ J}$.
14.
Why does the number of ejected photoelectrons increase with the intensity of incident light, whereas their kinetic energy remains independent of intensity?
Solution:
Intensity refers to the number of photons striking the surface per unit area per unit time. More photons mean more collisions with electrons, leading to a higher number of ejected photoelectrons (1 photon ejects 1 electron).
However, the Kinetic Energy depends solely on the energy of the individual photon ($E=h\nu$), which relies on frequency, not intensity.
15.
What is the radius of the first orbit of $\mathrm{He^+}$ ion? Compare it with the radius of the first orbit of Hydrogen atom.
Solution:
Bohr radius formula: $r_n = 0.529 \times \frac{n^2}{Z} \text{ Å}$.
For $\mathrm{He^+}$ ($Z=2, n=1$): $r_1 = 0.529 \times \frac{1^2}{2} = 0.2645 \text{ Å}$.
For H atom ($Z=1, n=1$): $r_1 = 0.529 \times \frac{1^2}{1} = 0.529 \text{ Å}$.
The radius of the first orbit of $\mathrm{He^+}$ is exactly half the radius of the first orbit of Hydrogen.
16.
Calculate the energy associated with the first orbit of $\mathrm{He^+}$. What is the radius of this orbit?
Solution:
Energy formula: $E_n = -2.18 \times 10^{-18} \times \left(\frac{Z^2}{n^2}\right) \text{ J}$.
For $\mathrm{He^+}$ ($Z=2, n=1$): $E_1 = -2.18 \times 10^{-18} \times \left(\frac{2^2}{1^2}\right) = -8.72 \times 10^{-18} \text{ J}$.
Radius (calculated in Q15) = $0.2645 \text{ Å}$ or $0.02645 \text{ nm}$.
17.
Calculate the wavelength of the spectral line obtained in the spectrum of $\mathrm{Li^{2+}}$ ion when the transition takes place between two levels whose principal quantum numbers are $n_2 = 3$ and $n_1 = 1$.
Solution:
Rydberg equation: $\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For $\mathrm{Li^{2+}}$, $Z = 3$. $n_1 = 1, n_2 = 3$. $R_H = 109677 \text{ cm}^{-1}$.
$\frac{1}{\lambda} = 109677 \times 3^2 \times \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 109677 \times 9 \times \left( \frac{8}{9} \right) = 109677 \times 8 = 877416 \text{ cm}^{-1}$.
$\lambda = \frac{1}{877416} \text{ cm} = 1.14 \times 10^{-6} \text{ cm} = 11.4 \text{ nm}$.
18.
What is the maximum number of emission lines when the excited electron of an H atom in $n = 6$ drops to the ground state?
Solution:
Formula for max number of spectral lines: $\frac{n(n-1)}{2}$ where $n$ is the excited state level.
Number of lines = $\frac{6(6-1)}{2} = \frac{6 \times 5}{2} = 15$ lines.
19.
Calculate the wavenumber for the shortest wavelength transition in the Balmer series of atomic hydrogen. (Given $R_H = 109677$ cm$^{-1}$)
Solution:
For the Balmer series, $n_1 = 2$.
Shortest wavelength means maximum energy, which corresponds to a transition from infinity ($n_2 = \infty$).
Wavenumber ($\bar{\nu}$) = $R_H \times 1^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{109677}{4} = 27419.25 \text{ cm}^{-1}$.
20.
Find the ratio of the velocity of an electron in the first Bohr orbit of Hydrogen to its velocity in the second Bohr orbit of $\mathrm{Li^{2+}}$.
Solution:
Velocity formula: $v_n \propto \frac{Z}{n}$.
For H ($n=1, Z=1$): $v_H \propto \frac{1}{1} = 1$.
For $\mathrm{Li^{2+}}$ ($n=2, Z=3$): $v_{Li} \propto \frac{3}{2} = 1.5$.
Ratio = $\frac{v_H}{v_{Li}} = \frac{1}{1.5} = \frac{2}{3}$ or $2:3$.
21.
The ionization energy of $\mathrm{He^+}$ is $19.6 \times 10^{-18}$ J atom$^{-1}$. Calculate the energy of the first stationary state of $\mathrm{Li^{2+}}$.
Solution:
Ionization energy (IE) is the energy required to remove an electron from the ground state: $IE = -E_1$.
$E_1 \propto Z^2$. Therefore, $\frac{IE(\mathrm{Li^{2+}})}{IE(\mathrm{He^+})} = \frac{Z_{Li}^2}{Z_{He}^2} = \frac{3^2}{2^2} = \frac{9}{4}$.
$IE(\mathrm{Li^{2+}}) = \frac{9}{4} \times 19.6 \times 10^{-18} = 44.1 \times 10^{-18} \text{ J}$.
Energy of first stationary state $E_1 = -IE = -44.1 \times 10^{-18} \text{ J atom}^{-1}$.
22.
If the velocity of an electron in a Bohr orbit is $V$, what will be its velocity in the next higher orbit of the same atom?
Solution:
Velocity is inversely proportional to the principal quantum number ($v \propto \frac{1}{n}$).
Let initial orbit be $n$, so $v_n = V$. The next orbit is $(n+1)$.
$\frac{v_{n+1}}{v_n} = \frac{n}{n+1} \Rightarrow v_{n+1} = V \left( \frac{n}{n+1} \right)$.
(e.g., if moving from $n=1$ to $n=2$, the velocity becomes $V/2$).
23.
Calculate the de Broglie wavelength of an electron traveling with a velocity equal to 1% of the speed of light.
Solution:
Velocity ($v$) = $1\% \text{ of } 3 \times 10^8 \text{ m/s} = 3 \times 10^6 \text{ m/s}$.
Mass of electron ($m$) = $9.1 \times 10^{-31} \text{ kg}$.
de Broglie wavelength ($\lambda$) = $\frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 3 \times 10^6}$
$\lambda \approx 2.42 \times 10^{-10} \text{ m} = 0.242 \text{ nm}$.
24.
A macroscopic ball of mass 0.1 kg is moving with a velocity of 10 m/s. Calculate its de Broglie wavelength and explain why the wave nature of this ball is not observable.
Solution:
$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{0.1 \times 10} = 6.626 \times 10^{-34} \text{ m}$.
This wavelength is incredibly small—far smaller than any measurable physical dimension or subatomic particle. Therefore, the wave character of macroscopic objects is negligible and completely unobservable.
25.
Two particles A and B are in motion. If the momentum of A is half of that of B, and if the wavelength of A is $5 \times 10^{-8}$ m, calculate the wavelength of B.
Solution:
$\lambda \propto \frac{1}{p}$ (where $p$ is momentum).
Given: $p_A = \frac{1}{2} p_B \Rightarrow \frac{p_A}{p_B} = \frac{1}{2}$.
$\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} = 2$.
$\lambda_B = \frac{\lambda_A}{2} = \frac{5 \times 10^{-8}}{2} = 2.5 \times 10^{-8} \text{ m}$.
26.
Calculate the kinetic energy of a moving electron which has a wavelength of 4.8 pm.
Solution:
$\lambda = 4.8 \text{ pm} = 4.8 \times 10^{-12} \text{ m}$.
Using relation: $K.E. = \frac{h^2}{2m\lambda^2}$
$K.E. = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (4.8 \times 10^{-12})^2}$
$K.E. = \frac{43.90 \times 10^{-68}}{18.2 \times 10^{-31} \times 23.04 \times 10^{-24}} = \frac{43.90 \times 10^{-68}}{419.3 \times 10^{-55}} \approx 1.047 \times 10^{-14} \text{ J}$.
27.
Derive the relationship between the de Broglie wavelength ($\lambda$) and the Kinetic Energy (K.E.) of a particle.
Solution:
Kinetic Energy ($K.E.$) = $\frac{1}{2}mv^2$. Multiply and divide by $m$:
$K.E. = \frac{(mv)^2}{2m} = \frac{p^2}{2m}$, where $p$ is momentum.
Solving for $p$: $p = \sqrt{2m(K.E.)}$.
From de Broglie's equation: $\lambda = \frac{h}{p}$.
Substituting $p$: $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.
28.
A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.
Solution:
Mass ($m$) = $40 \text{ g} = 0.04 \text{ kg}$. Velocity ($v$) = $45 \text{ m/s}$.
Uncertainty in speed ($\Delta v$) = $2\% \text{ of } 45 = \frac{2}{100} \times 45 = 0.9 \text{ m/s}$.
Heisenberg equation: $\Delta x \cdot m\Delta v = \frac{h}{4\pi}$
$\Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 0.04 \times 0.9} = \frac{6.626 \times 10^{-34}}{0.4524} \approx 1.46 \times 10^{-33} \text{ m}$.
29.
If the position of an electron is measured with an accuracy of $\pm 0.002$ nm, calculate the uncertainty in the momentum of the electron.
Solution:
Uncertainty in position ($\Delta x$) = $0.002 \text{ nm} = 2 \times 10^{-12} \text{ m}$.
$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$
$\Delta p = \frac{h}{4\pi \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 2 \times 10^{-12}}$
$\Delta p \approx 2.63 \times 10^{-23} \text{ kg m/s}$.
30.
Show that if the uncertainty in the position of a moving particle is equal to its de Broglie wavelength, its velocity is completely uncertain (i.e., $\Delta v \approx v/4\pi$).
Solution:
Given $\Delta x = \lambda$. We know $\lambda = \frac{h}{mv}$. So, $\Delta x = \frac{h}{mv}$.
Substitute $\Delta x$ into Heisenberg's equation ($\Delta x \cdot m\Delta v = \frac{h}{4\pi}$):
$\left(\frac{h}{mv}\right) \cdot m\Delta v = \frac{h}{4\pi}$
$\frac{\Delta v}{v} = \frac{1}{4\pi} \implies \Delta v = \frac{v}{4\pi}$.
31.
Why is Heisenberg's uncertainty principle significant only for microscopic particles and negligible for macroscopic objects? Provide mathematical reasoning.
Solution:
According to the principle, $\Delta x \cdot \Delta v = \frac{h}{4\pi m}$.
The term $\frac{h}{4\pi}$ is extremely small ($\sim 10^{-34}$). For a macroscopic object, the mass $m$ is relatively large (e.g., 1 kg), making the product $\Delta x \cdot \Delta v \approx 10^{-34}$, which is unmeasurable and practically zero.
For an electron, mass $m \approx 10^{-31}$ kg, making $\Delta x \cdot \Delta v \approx 10^{-4}$, which is highly significant and measurable.
32.
List all the possible values of $l$ and $m_l$ for $n = 3$. How many total orbitals are present in the 3rd shell?
Solution:
For $n = 3$, $l$ can take values from $0$ to $(n-1)$. So, $l = 0, 1, 2$.
If $l = 0$ (3s), $m_l = 0$ (1 orbital).
If $l = 1$ (3p), $m_l = -1, 0, +1$ (3 orbitals).
If $l = 2$ (3d), $m_l = -2, -1, 0, +1, +2$ (5 orbitals).
Total orbitals = $1 + 3 + 5 = 9$. (Formula: $n^2 = 3^2 = 9$).
33.
Designate the orbitals having: (a) $n=2, l=1$ (b) $n=4, l=0$ (c) $n=5, l=3$ (d) $n=3, l=2$.
Solution:
(a) 2p
(b) 4s
(c) 5f ($l=0 \rightarrow s, l=1 \rightarrow p, l=2 \rightarrow d, l=3 \rightarrow f$)
(d) 3d
34.
Calculate the total number of angular nodes and radial nodes present in a 4d orbital.
Solution:
For a 4d orbital, $n = 4$ and $l = 2$.
Angular nodes = $l = 2$.
Radial nodes = $n - l - 1 = 4 - 2 - 1 = 1$.
(Total nodes = $n - 1 = 3$).
35.
What is the maximum number of electrons in an atom that can have the quantum numbers $n=4, m_s=-1/2$?
Solution:
Total number of electrons in a shell $n$ is $2n^2$. For $n=4$, max electrons = $2(4)^2 = 32$.
According to Pauli's Exclusion Principle, exactly half of these will have spin $m_s = +1/2$ and the other half will have $m_s = -1/2$.
Max electrons with $m_s = -1/2$ is $32 / 2 = 16$.
36.
Write the formula for orbital angular momentum. Calculate its value for a 3p electron.
Solution:
Formula: $\mu_L = \sqrt{l(l+1)} \frac{h}{2\pi}$.
For a 3p electron, the azimuthal quantum number $l = 1$.
$\mu_L = \sqrt{1(1+1)} \frac{h}{2\pi} = \sqrt{2} \frac{h}{2\pi}$ (or $\sqrt{2}\hbar$).
37.
Identify which of the following sets of quantum numbers are not possible and state the reason: (a) $n=1, l=0, m_l=0, m_s=-1/2$ (b) $n=3, l=3, m_l=-3, m_s=+1/2$ (c) $n=2, l=1, m_l=0, m_s=+1$.
Solution:
(a) Possible (represents a 1s electron).
(b) Not possible. The value of $l$ must be less than $n$ ($l = 0$ to $n-1$). Here, $n=3$ and $l=3$ is invalid.
(c) Not possible. The spin quantum number $m_s$ can only take values $+1/2$ or $-1/2$. $m_s = +1$ is invalid.
38.
State Pauli's Exclusion Principle. Explain how it determines the maximum capacity of an orbital to hold electrons.
Solution:
Statement: No two electrons in an atom can have the same set of all four quantum numbers ($n, l, m_l, m_s$).
Explanation: An orbital is defined by three quantum numbers ($n, l, m_l$). If two electrons occupy the same orbital, they share these three. To satisfy the principle, they must have different spin quantum numbers ($m_s$). Since $m_s$ only has two possible values ($+1/2, -1/2$), an orbital can hold a maximum of 2 electrons, and they must have opposite spins.
39.
Write the complete electronic configuration of $\mathrm{Cr}$ (Z=24) and $\mathrm{Cu}$ (Z=29). Explain the reason for their exceptional configurations.
Solution:
Cr (24): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$
Cu (29): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}$
Reason: Moving one electron from the 4s to the 3d orbital creates a completely half-filled ($3d^5$) or fully-filled ($3d^{10}$) subshell. These configurations are exceptionally stable due to (a) symmetrical distribution of electron charge, and (b) maximum exchange energy between parallel spins.
40.
Determine the number of unpaired electrons in the following ions: (a) $\mathrm{Fe^{3+}}$ (Z=26) (b) $\mathrm{Mn^{2+}}$ (Z=25) (c) $\mathrm{Ni^{2+}}$ (Z=28).
Solution:
(a) $\mathrm{Fe}$ is $[Ar] 4s^2 3d^6$. $\mathrm{Fe^{3+}}$ is $[Ar] 3d^5$. There are 5 unpaired electrons.
(b) $\mathrm{Mn}$ is $[Ar] 4s^2 3d^5$. $\mathrm{Mn^{2+}}$ is $[Ar] 3d^5$. There are 5 unpaired electrons.
(c) $\mathrm{Ni}$ is $[Ar] 4s^2 3d^8$. $\mathrm{Ni^{2+}}$ is $[Ar] 3d^8$. Filling 8 electrons in 5 d-orbitals leaves 2 unpaired electrons.
41.
Based on Hund's Rule of Maximum Multiplicity, write the orbital diagram configuration for Nitrogen (Z=7) and Oxygen (Z=8).
Solution:
Hund's rule states electron pairing in degenerate orbitals won't occur until each orbital is singly occupied with parallel spins.
Nitrogen (7): $1s^2 2s^2 2p^3$. The 2p orbitals are: $2p_x^1 \ (\uparrow)$, $2p_y^1 \ (\uparrow)$, $2p_z^1 \ (\uparrow)$.
Oxygen (8): $1s^2 2s^2 2p^4$. The 2p orbitals are: $2p_x^2 \ (\uparrow\downarrow)$, $2p_y^1 \ (\uparrow)$, $2p_z^1 \ (\uparrow)$.
42.
Calculate the spin-only magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. (Formula: $\mu = \sqrt{n(n+2)}$ BM)
Solution:
Atomic number Z=25 corresponds to Manganese (Mn).
Divalent ion is $\mathrm{Mn^{2+}}$. Its configuration is $[Ar] 3d^5$.
Number of unpaired electrons ($n$) = 5.
Magnetic moment ($\mu$) = $\sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ Bohr Magnetons (BM)}$.
43.
Arrange the following orbitals in increasing order of their energy: 4d, 5p, 5s, 6s. (Use $(n+l)$ rule).
Solution:
$(n+l)$ values:
- 4d: $n=4, l=2 \Rightarrow (n+l) = 6$
- 5p: $n=5, l=1 \Rightarrow (n+l) = 6$
- 5s: $n=5, l=0 \Rightarrow (n+l) = 5$
- 6s: $n=6, l=0 \Rightarrow (n+l) = 6$
Lower $(n+l)$ means lower energy. For orbitals with the same $(n+l)$, lower $n$ means lower energy.
Order: $5s \ (\text{sum=5}) < 4d \ (\text{sum=6, } n=4) < 5p \ (\text{sum=6, } n=5) < 6s \ (\text{sum=6, } n=6)$.
